difflib 是python提供的比较序列(string list)差异的模块。
import difflib from pprint import pprint a = 'pythonclub.org is wonderful' b = 'Pythonclub.org also wonderful' #构造SequenceMatcher类 s = difflib.SequenceMatcher(None, a, b) #得到相同的block print "s.get_matching_blocks():" pprint(s.get_matching_blocks()) print print "s.get_opcodes():" for tag, i1, i2, j1, j2 in s.get_opcodes(): print ("%7s a[%d:%d] (%s) b[%d:%d] (%s)" % (tag, i1, i2, a[i1:i2], j1, j2, b[j1:j2])) #在此实现你的功能
s.get_matching_blocks(): [(1, 1, 14), (16, 17, 1), (17, 19, 10), (27, 29, 0)] s.get_opcodes(): replace a[0:1] (p) b[0:1] (P) equal a[1:15] (ythonclub.org ) b[1:15] (ythonclub.org ) replace a[15:16] (i) b[15:17] (al) equal a[16:17] (s) b[17:18] (s) insert a[17:17] () b[18:19] (o) equal a[17:27] ( wonderful) b[19:29] ( wonderful)
import difflib str1 = "Poor Impulse Control: A Good Babysitter Is Hard To Find" str2 = """ A Good Babysitter Is Hard To Find This is Frederick by Leo Lionni, the first book I picked for myself. I was in kindergarten, I believe, which would be either 1968 or 1969. Frederick has a specific lesson for children about how art is as important in life as bread, but there's a secondary consideration I took away: if we pool our talents our lives are immeasurably better. Curiously, this book is the story of my life, however one interprets those things. I expect Mickey Rooney to show up any time with a barn and a plan for a show, though my mom is not making costumes. My sisters own a toy store with a fantastic selection of imaginative children's books. I try not to open them because I can't close them and put them back. My tantrums are setting a bad example for the kids. Anyway, I mention this because yesterday was Mr. Rogers' 40th anniversary. I appreciate the peaceful gentleman more as time passes, as I play with finger puppets in department meetings, as I eye hollow trees for Lady Elaine Fairchild infestations. Maybe Pete can build me trolley tracks!Labels: To Take Your Heart Away """ s = difflib.SequenceMatcher(None, str1, str2) print len(str1), len(str2) star_a, start_b, length = s.find_longest_match(0, len(str1)-1, 0, len(str2)-1) print star_a, start_b, length print str1[star_a:star_a + length]
55 1116 0 1048 1 P 版本为: Python 2.5.1 (r251:54863, Apr 18 2007, 08:51:08) [MSC v.1310 32 bit (Intel)] on win32 Type "help", "copyright", "credits" or "license" for more information. >>>
而最长的应该为 A Good Babysitter Is Hard To Find.
将 str1 于 str2 交换一下， len(str1) > len(str2).
Work Around为: 将较长的字符串设为第一个，而较短的设为第二个。